Problem: Simplify the following expression: $n = \dfrac{-4x^2 + 16x + 128}{x - 8} $
First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $-4$ , so we can rewrite the expression: $ n =\dfrac{-4(x^2 - 4x - 32)}{x - 8} $ Then we factor the remaining polynomial: $x^2 {-4}x {-32} $ ${-8} + {4} = {-4}$ ${-8} \times {4} = {-32}$ $ (x {-8}) (x + {4}) $ This gives us a factored expression: $\dfrac{-4(x {-8}) (x + {4})}{x - 8}$ We can divide the numerator and denominator by $(x + 8)$ on condition that $x \neq 8$ Therefore $n = -4(x + 4); x \neq 8$